Q 12-11-146JEE MainJEE Main 2020 (7 Jan, Shift 2)Medium
An electron (of mass $m$) and a photon have the same energy $E$ in the range of a few eV. The ratio of the de Broglie wavelength associated with the electron and the wavelength of the photon is ($c$ = speed of light in vacuum):
Answer: (C) $\dfrac1c\left(\dfrac{E}{2m}\right)^{1/2}$
Electron (non-relativistic at a few eV): $\lambda_e = \dfrac{h}{\sqrt{2mE}}$. Photon: $\lambda_p = \dfrac{hc}{E}$.
$$\frac{\lambda_e}{\lambda_p} = \frac{E}{c\sqrt{2mE}} = \frac1c\left(\frac{E}{2m}\right)^{1/2}$$
(The official answer key lists $\left(\frac{E}{2m}\right)^{1/2}$, which has the dimension of speed and cannot be a ratio of two wavelengths.)
Solution by Sreeraj P, M.Sc Physics